You are correct that if no enzyme is present, the petals will remain white. The interesting part of this question is that there are two enzymes, each catalyzing a step in the biosynthetic pathway for the blue pigment, and that the biosynthetic pathway has a different-colored (lilac) intermediate.

The electrophoresis pattern shown in the top row reveals a single enzyme present and corresponds to white petals observed. If the band was from enzyme 1, the enzyme would have catalyzed the transformation of the uncolored pigment precursor to its lilac form. However, the petals remained white, so we can conclude that the band reveals enzyme 2 and enzyme 1 is missing; if enzyme 1 were present it would have made the lilac pigment from the uncolored precursor but, without enzyme 1 or the lilac pigment present, enzyme 2 can be present in the cell but cannot produce the blue pigment because its substrate (lilac pigment) is not present.

In the second row, both enzymes are present (both bands appeared in the electrophoretic lane). This means enzyme 1 is present and can catalyze the formation of the lilac pigment from its uncolored precursor. Enzyme 2 is also present, so the lilac pigment is transformed into the blue pigment and the petals appear blue.

In the third row, only enzyme 1 is present. It catalyzes the transformation of the uncolored precursor into the lilac pigment. Because enzyme 2 is not present to use up the lilac pigment, that pigment accumulates and gives the petals their lilac color.

In the fourth row, with both enzymes missing from the electrophoretic lane, the uncolored precursor is not changed into colored forms and the petals remain white.

This gives rise to the colors in the answers you reported:R1 (white)R2 blueR3 lilacR4 white